Solution to puzzle 116: Factorial divisors
Show that, for each n 
3, n! can be represented as the sum of n distinct divisors of itself. (For example, 3! = 1 + 2 + 3.)
We will use mathematical induction on n to prove the slightly stronger result that, for each n 3, n! can be represented as the sum of n distinct divisors of itself, with 1 as one of the divisors..
We have the base case, for n = 3: 3! = 1 + 2 + 3, with 1 as one of the divisors.
For the induction step, we assume the inductive hypothesis, that we have a representation for n = k, with 1 as one of the divisors.
That is, k! = 1 + d2 + d3 + ... + dk, with 1 < d2 < ... < dk.
| Then (k + 1)! | = (k + 1) + (k + 1)d2 + ... + (k + 1)dk |
| = 1 + k + (k + 1)d2 + ... + (k + 1)dk |
That is, the sum of k + 1 distinct divisors of (k + 1)!, with 1 as one of the divisors.
It follows by induction that, for each n 3, n! can be represented as the sum of n distinct divisors of itself.
A generalization 
Show that any given positive integer less than or equal to n! can be represented as the sum of at most n distinct divisors of n!.
Hint - Solution
Source: Mathematical Miniatures, by Svetoslav Savchev and Titu Andreescu. See Chapter 15.